Douglas C. Giancoli Chapter 21 - Physics & Astronomy
Transcript of Douglas C. Giancoli Chapter 21 - Physics & Astronomy
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Chapter 21
Electric Charge andElectric Field
Physics for Scientists & Engineers, 3rd EditionDouglas C. Giancoli
© Prentice Hall
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7P01 -
Why Study Physics?
• Understand/Appreciate Nature• Understand Technology• Learn to Solve Difficult Problems• It’s Required
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19P01 -
Physics is not Math…
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20P01 -
…but we use concepts from 18.02
V= −∇E•Gradients
•Path Integrals
•Surface Integrals
•Volume Integrals
B
AV d∆ ≡ − ⋅∫ E s
0
in
S
Qdε
⋅ =∫∫ E A
Q dVρ= ∫∫∫
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25P01 -
8.02: Electricity and MagnetismAlso new way of thinking…
How do objects interact at a distance? Fields We will learn about E & M Fields:
how they are created & what they effectBig Picture Summary:
0
0 0 00
in B
S C
Eenc
S C
Q dd ddt
dd d Idt
ε
µ µ ε
Φ⋅ = ⋅ = −
Φ⋅ = ⋅ = +
∫∫ ∫
∫∫ ∫
E A E s
B A B s
MaxwellEquations:
( )q= + ×F E v BLorentz Force:
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46P01 -
Electric Charge (~Mass)Two types of electric charge: positive and negativeUnit of charge is the coulomb [C]
Charge of electron (negative) or proton (positive) is
Charge is quantized
Charge is conserved
19, 1.602 10e e C−± = ×
Q Ne= ±
n p e ν−→ + + e e γ γ+ −+ → +
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Figure 21-3
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Figure 21-4
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Figure 21-5 (a)
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Figure 21-5 (b)
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Figure 21-5 (c)
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Figure 21-2
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Figure 21-6
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47P01 -
Electric Force (~Gravity)The electric force between charges q1 and q2 is
(a) repulsive if charges have same signs (b) attractive if charges have opposite signs
Like charges repel and opposites attract !!
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Figure 21-14
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48P01 -
Coulomb's Law
1 212 2
ˆeq qkr
=F rCoulomb’s Law: Force by q1 on q2
9 2 2
0
1 8.9875 10 N m /C4ekπε
= = ×
ˆ :r unit vector from q1 to q2
rrr =ˆ 1 2
12 3eq qkr
⇒ =F r
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Figure 21-15
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Figure 21-16
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50P01 -
The Superposition PrincipleMany Charges Present:Net force on any charge is vector sum of forces from other individual charges
3 13 23= +F F FExample:
In general:
1
N
j iji=
=∑F F
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Figure 21-17
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Figure 21-18
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49P01 -
Coulomb's Law: Example
( )3132 2 2
ˆ ˆ m
1mr
= −
=
r i j?32 =F
a = 1 m
q1 = 6 C
q3 = 3 C
q2 = 3 C
32r
( )( )( )( )( )
129 2 2
3
ˆ ˆ3 m9 10 N m C 3C 3C
1m
−= ×
i j32 3 2 3ek q q
r=
rF
( )981 10 ˆ ˆ3 N
2×
= −i j
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Figure 21-19
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Figure 21-20
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Figure 21-32
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Figure 21-33 (a)
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Figure 21-33 (b)
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35P01 -
Visualizing Vector Fields:Three Methods
Vector Field DiagramArrows (different colors or length) in direction of field
on uniform grid.Field Lines
Lines tangent to field at every point along lineGrass Seeds
Textures with streaks parallel to field direction
All methods illustrated inhttp://ocw.mit.edu/ans7870/8/8.02T/f04/visualization
s/electrostatics/39-pcharges/39-twocharges320.html
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36P01 -
Vector Fields – Field Lines
• Direction of field line at any point is tangent to field at that point
• Field lines never cross each other
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38P01 -
Vector Fields – “Grass Seeds”
Source/Sink Circulating
Although we don’t know absolute direction, we can determine relative direction
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42P01 -
2ˆg
MmGr
= −F r
Example Of Vector Field: Gravitation
Gravitational Force:
Gravitational Field:
2
2
/ ˆ ˆg GMm r MGm m r
= = − = −F
g r r
M : Mass of Earth
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43P01 -
Example Of Vector Field: Gravitation
Gravitational Field:
2ˆMG
r= −g r g m=F g
Created by M Felt by m
ˆ :r unit vector from M to m
rrr =ˆ
M : Mass of Earth
3
MGr
⇒ = −g r
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45P01 -
From Gravitational toElectric Fields
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51P01 -
Electric Field (~g)The electric field at a point is the force acting on a test charge q0 at that point, divided by the charge q0 :
0q≡
FE
2ˆe
qkr
=E rFor a point charge q:
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/04-MovingChargePosElec/04-MovChrgPosElec_f223_320.html
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52P01 -
Superposition Principle
The electric field due to a collection of N point charges is the vector sum of the individual electric fields due to each charge
1 21
.. . . .N
total ii=
= + + =∑E E E E
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53P01 -
Summary Thus Far
Mass M Charge q (±)
2ˆMG
r= −g r 2
ˆeqkr
=E rCREATE:
g m=F g E q=F E
This is easiest way to picture field
FEEL:
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Figure 21-33 (c)
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Figure 21-21
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Figure 21-23
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Figure 21-25
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Figure 21-26
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6P02 -
In-Class Problem
d
s
q− q+
P
ij
Consider two point charges of equal magnitude but opposite signs, separated by a distance d. Point Plies along the perpendicular bisector of the line joining the charges, a distance s above that line. What is the E field at P?
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24P02 -
Continuous Charge Distributions
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25P02 -
Continuous Charge Distributions
( ) ?P =E
Vi
iQ q
Break distribution into parts:
= ∆∑
2ˆe
qkr∆
∆ =E r
E field at P due to ∆q
Superposition:
= ∆∑E E
V
dq→ ∫
d→ ∫ E
2ˆe
dqd kr
→ =E r
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26P02 -
Continuous Sources: Charge DensitydVdQ ρ=
R
L
2Volume V R Lπ= = QV
ρ =
LQ
=λ
QA
σ =
dAdQ σ=w
L
Area A wL= =
dLdQ λ=Length L=
L
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27P02 -
Examples of Continuous Sources: Line of charge
LQ
=λ
Length L=
L
dLdQ λ=
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect
rostatics/07-LineIntegration/07-
LineInt320.html
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28P02 -
Examples of Continuous Sources: Line of charge
LQ
=λ
Length L=
L
dLdQ λ=
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect
rostatics/08-LineField/08-
LineField320.html
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36P02 -
r
2L
−2L
+
s
P
j
i
In-Class: Line of Charge
Point P lies on perpendicular bisector of uniformly charged line of length L, a distance s away. The charge on the line is Q. What is E at P?
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37P02 -
r
θ
θ
2L
−2L
+
xd ′
x′
xddq ′= λs
22 xsr ′+=
P
j
i
Hint: http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/07-LineIntegration/07-LineInt320.html
Typically give the integration variable (x’) a “primed” variable name.
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32P02 -
Ring of Charge
Symmetry!0E⊥ =1) Think about it
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect
rostatics/09-RingIntegration/09-
ringInt320.html
2) Define Variables
dq dlλ=22 xar +=
( )a dλ ϕ=
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38P02 -
E Field from Line of Charge
2 2 1/ 2ˆ
( / 4)eQk
s s L=
+E j
Limits:
2ˆlim e
s L
Qks>>
→E j Point charge
ˆ ˆ2 2lim e es L
Qk kLs s
λ<<
→ =E j j Infinite charged line
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29P02 -
Examples of Continuous Sources: Ring of Charge
2Q
Rλ
π=dLdQ λ=
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect
rostatics/09-RingIntegration/09-
ringInt320.html
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30P02 -
Examples of Continuous Sources: Ring of Charge
2Q
Rλ
π=dLdQ λ=
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect
rostatics/10-RingField/10-
ringField320.html
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31P02 -
Example: Ring of Charge
P on axis of ring of charge, x from centerRadius a, charge density λ.
Find E at P
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Figure 21-27
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33P02 -
Ring of Charge3) Write Equation dq a dλ ϕ=
2
ˆe
rd k dqr
=E
a) My way
3x exdE k dqr
=
b) Another way
22 xar +=
3erk dqr
=
cos( )xdE d θ= E 2 3
1e e
x xk dq k dqr r r
= ⋅ =
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34P02 -
Ring of Charge4) Integrate
3x x exE dE k dqr
= =∫ ∫22 xar +=
dq a dλ ϕ=
3exk dqr
= ∫
Very special case: everything except dq is constant
2aλ π=dq∫2 2
0 0a d a d
π πλ ϕ λ ϕ= =∫ ∫
Q=
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35P02 -
Ring of Charge5) Clean Up
3x exE k Qr
=
( )3/ 22 2x exE k Q
a x=
+
0a →
( )3/ 22 2ˆ
exk Q
a x=
+E i
6) Check Limit
( )3/ 2 22
ex e
k QxE k Qxx
→ =
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39P02 -
In-Class: Uniformly Charged Disk
P on axis of disk of charge, x from centerRadius R, charge density σ.
Find E at P
( 0 )x >
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40P02 -
Disk: Two Important Limits
( )1/ 22 2ˆ1
2disko
x
x R
⎡ ⎤⎢ ⎥= −⎢ ⎥+⎣ ⎦
E iσε
Limits:
2
1 ˆlim 4diskx R o
Qx>>
→E iπε
*** Point charge
ˆlim 2diskx R o<<
→E iσε
Infinite charged plane
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41P02 -
E for Plane is Constant????
1) Dipole: E falls off like 1/r3
2) Point charge: E falls off like 1/r2
3) Line of charge:E falls off like 1/r4) Plane of charge: E constant
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Figure 21-30
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Figure 21-33 (d)
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Figure 21-35
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Figure 21-36
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Figure 21-37
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Figure 21-39
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Figure 21-40
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12P02 -
Electric DipoleTwo equal but opposite charges +q and –q,
separated by a distance 2a
q
-q
2a charge×displacementˆ ˆ×2 2q a qa
≡
= =
p
j j
Dipole Moment
p
p points from negative to positive charge
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Figure 21-41
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13P02 -
Why Dipoles?
Nature Likes To Make Dipoles!
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/20-Molecules2d/20-mole2d320.html
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14P02 -
Dipoles make Fields
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15P02 -
Electric Field Created by DipoleThou shalt use components!
3 3x ex xE k q
r r+ −
⎛ ⎞∆ ∆= −⎜ ⎟
⎝ ⎠
3 3y ey yE k q
r r+ −
+ −
⎛ ⎞∆ ∆= −⎜ ⎟
⎝ ⎠
3/2 3/22 2 2 2( ) ( )e
x xk qx y a x y a
⎛ ⎞⎜ ⎟= −⎜ ⎟⎡ ⎤ ⎡ ⎤+ − + +⎣ ⎦ ⎣ ⎦⎝ ⎠
3/2 3/22 2 2 2( ) ( )e
y a y ak qx y a x y a
⎛ ⎞− +⎜ ⎟= −⎜ ⎟⎡ ⎤ ⎡ ⎤+ − + +⎣ ⎦ ⎣ ⎦⎝ ⎠
2 3 3 3
ˆ ˆ ˆx yr r r r
∆ ∆= = +
r r i j
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19P02 -
Dipoles feel Fields
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21P02 -
Dipole in Uniform FieldˆE=E i
ˆ ˆ2 (cos sin )qa= +p i jθ θ
( ) 0net q q+ −= + = + − =F F F E E
tends to align with the electric field p
Total Net Force:
Torque on Dipole: = ×τ r F( )( )2 sin( )a qE θ=
= ×p Esin( )rFτ θ+= sin( )pE θ=
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Figure 21-43
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Figure 21-47
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Figure 21-50
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Figure 21-51
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Figure 21-52
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Figure 21-53
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Figure 21-55
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Figure 21-56
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Figure 21-57
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Figure 21-59
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Figure 21-60
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Figure 21-61
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Figure 21-62
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Figure 21-63
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Figure 21-66
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Figure 21-67
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Figure 21-68
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Figure 21-69
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Figure 21-70
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Figure 21-71
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Figure 21-72